Question

( those who will answer I will mark as brainliest) Line ‘l’ is the bisector of ∠A and B is any point on ‘l’. BP and BQ are perpendiculars from B to the arms of ∠A. Show that i) ∆ APB ≅ ∆ AQB and ii) B is equidistant from the arms of ∠A. (or BP = BQ)

Answer 1

Answer:

Given line 'l' is the bisector of Angle A and B is any point on L. BP and BQ are perpendiculars from B to the arms of ∠A.

Step-by-step explanation:

IN TRIANGLE APB AND AQB

ANGLE BAP= BAQ

AB= AB( COMMON)

ANGLE BPA=BQA

TRIANGLE APB IS CONGRUENT TO TRIANGLE AQB (AAS)

BP=BQ(CPCT)

Answer 2

Answer:

Given: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. B is equidistant from the arms of ∠A. ... To Prove: ∠BCD is a right angle.